RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

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# RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:35 PM IST

By regularly practicing RD Sharma Class 12th Exercise 17.3, understudies will be severe with the thoughts given in the Class 12 RD Sharma textbook. Be that as it may, it is nearly impossible to be skilled at maths who thought proper practice. The Rd Sharma class 12th exercise 17.3 will help students to learn and practice at home so that they have the chance to improve their skills to score well in boards and JEE mains.

## Maxima and Minima Excercise: 17.3

Maxima and Minima excercise 17.3 question 1 (i)

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) < 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value
Given:
$f(x)=x^{4}-62 x^{2}+120 x+9$
Explanation:
It is given that
\begin{aligned} &f(x)=x^{4}-62 x^{2}+120 x+9 \\ &f^{\prime}(x)=4 x^{3}-62 * 2 x+120 \end{aligned}
To find critical values ,
\begin{aligned} &f^{\prime}(x)=0 \\ &4 x^{3}-124 x+120=0 \\ &4\left(x^{3}-31 x+30\right)=0 \end{aligned}
On solving,
critical values of x are 5,1,-6.
Now,
$f^{\prime \prime}(x)=12 x^{2}-1$
At x=5,
\begin{aligned} f^{\prime \prime}(5) &=12(5)^{2}-124 \\ &=176>0 \end{aligned}
So, x=5 is point of local minima.
At x=1,
\begin{aligned} f^{\prime \prime}(1) &=12(1)^{2}-124 \\ &=-112<0 \end{aligned}
So, x=1 is point of local maxima.
At x=-6,
\begin{aligned} f^{\prime \prime}(-6) &=12(-6)^{2}-124 \\ &=308>0 \end{aligned}

So, x=-6 is a point of local minima.
The local max. value at x=1 is
\begin{aligned} f(1) &=(1)^{4}-62(1)^{2}+120(1)+9 \\ &=68 \end{aligned}
And local min. value at x=5 & x=-6 are,
\begin{aligned} f(5)=&(5)^{4}-62(5)^{2}+120(5)+9 \\ &=-316 \\ f(-6) &=(-6)^{4}-62(-6)^{2}+120(-6)+9 \\ &=-1647 \end{aligned}
Hence, point of local maxima is 1 & its maximum value is 68.
Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-6 x^{2}+9 x+15$
Explanation:
We have,

\begin{aligned} f(x) &=x^{3}-6 x^{2}+9 x+15 \\ f^{\prime}(x) &=3 x^{2}-6.2 x+9 \\ &=3 x^{2}-12 x+9 \\ \therefore f^{\prime}(x) &=3\left(x^{2}-4 x+3\right) \\ \& f^{\prime \prime}(x) &=3(2 x-4) \\ &=6 x-12 \end{aligned}
To find maxima and minima.
\begin{aligned} &\qquad f^{\prime}(x)=0 \\ &3\left(x^{2}-4 x+3\right)=0 \\ &\quad x^{2}-4 x+3=0 \\ &X=3 \text { or } x=1 \\ &\text { At } x=3, \\ &f^{\prime \prime}(3)=6(3)-12 \\ &=6>0 \end{aligned}
At x = 3,
f’’(3) = 6(3)-12
=6>0
So, x = 3 is point of local minima
At x = 1,
f’’(1) = 6(1)-12
=-6<0
So, x = 1 is point of local maxima
Now, local maximum value at x= 1 is
\begin{aligned} &f(1)=(1)^{3}-6(1)^{2}+9(1)+15 \\ &f(1)=19 \end{aligned}

And local min. value at x=3 is
\begin{aligned} &f(3)=(3)^{3}-6(3)^{2}+9(3)+15 \\ &f(3)=15 \end{aligned}
Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

## Maxima and Minima excercise 17.3 question 1 (iii)

Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is -4
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x-1)(x+2)^{2}$
Explanation:
We have,
\begin{aligned} f(x) &=(x-1)(x+2)^{2} \\ f^{\prime}(x) &=(x+2)^{2}(1)+(x-1) 2(x+2) \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=(x+2)(x+2+2 x-2) \\ &=(x+2)(3 x) \\ f^{\prime \prime}(x) &=3 x(1)+(x+2) 3 \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=6 x+6 \end{aligned}
For max and min, f’(x)=0,
\begin{aligned} &(x+2)(3 x)=0 \\ &x=0 \& x=-2 \end{aligned}
At x=0,
$f^{\prime \prime}(0)=6(0)+6=6>0$
So, x= 0 is point of local minima
$f^{\prime \prime}(0)=6(0)+6=6>0$
At x=-2,
\begin{aligned} f^{\prime \prime}(0) &=6(-2)+6 \\ &=-6<0 \end{aligned}
So, x = -2 is point of local maxima
So, local max. value at x=-2 is
f(-2)=(-2-1)(-2+2)^{2}=0
And local min. value at x= 0 is
$f(0)=(0-1)(0+2)^{2}=-4$
Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.

Point of local maxima is 2 and it’s local maximum value is $\frac{1}{2}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f"(c_2) <0$then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0$
Explanation:
We have,
$\begin{gathered} f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0 \\ \therefore f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{4}{x^{3}} \\ f^{\prime \prime}(x)=\frac{4}{x^{2}}-\frac{12}{x^{4}} \end{gathered}$

For maxima and minima, f’(x)=0
$\begin{array}{r} -\frac{2}{x^{2}}+\frac{4}{x^{3}}=0 \\ -\frac{2(x-2)}{x^{3}}=0 \\ x=2 \end{array}$
At x=2,
\begin{aligned} f^{\prime \prime}(2) &=\frac{4}{(2)^{3}}-\frac{12}{(2)^{2}} \\ &=\frac{4}{8}-\frac{12}{16} \\ &=\frac{1}{2}-\frac{3}{4} \\ &=-\frac{1}{4}<0 \end{aligned}
So, x=2 is point of local maxima.
So, local max. value at x= 2 is
\begin{aligned} f(2) &=\frac{2}{2}-\frac{2}{(2)^{2}} \\ &=1-\frac{2}{4}=\frac{1}{2} \end{aligned}
Hence, point of local maxima is 2 & its max. value is $\frac{1}{2}$

## Maxima and Minima Excercise 17.3 question 1 (v)

Point of local minima value is x=-1 and it’s local minimum value is $-\frac{1}{e^{.}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) > 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:$f(x)=x e^{\mathrm{x}}$
Explanation:
We have,
\begin{aligned} &f(x)=x e^{\mathrm{x}} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \ldots \text { using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+1)+e^{\mathrm{x}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+2) \end{aligned}
Now, for maxima & minima, f’(x)=0
\begin{aligned} e^{\mathrm{x}}(x+1) &=0 \\ x+1 &=0 \\ x &=-1 \end{aligned}
at x=-1,
$f^{\prime \prime}(-1)=e^{-1}(-1+2)=\frac{1}{e}$
X = -1 is point of local minima.
So, local min. value at x= 1 is
$f(-1)=-1 e^{-1}=-1 / e$
Thus, point of local minima is -1 & its min value is $-\frac{1}{e}$

## Maxima and Minima exercise 17.3 question 1 (vi)

Point of local minima value is 2 and it’s local minimum value is 2.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $\mathrm{c}_{1}$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$\frac{x}{2}+\frac{2}{x}, \underline{x}>0$
Explanation:
We have,
\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned}
For local maxima or minima, we have $f^{\prime}(x)=0$
$\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}$
since, $x>0$
$x = 2$
Now, at x=2 ,
$f(2)=\frac{2}{2}+\frac{2}{2}=2$
Thus, its min. value at 2 is 2.

Point of local minima value is $-\frac{7}{4}$and it’s local minimum value is $-\frac{3}{(4)^{\frac{4}{3}}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x+1)(x+2)^{\frac{1}{3}}, x>=-2$
Explanation:
we have ,
\begin{aligned} &f(x)=(x+1)(x+2)^{-} \\ &f^{\prime}(x)=(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{\left(-\frac{2}{3}\right)} \ldots \text{ using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=\frac{2}{3(x+2)^{2 / 3}}-\frac{2}{9}(x+1)(x+2)^{\left(-\frac{5}{3}\right)} \ldots \text{ using }\frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \end{aligned}
For maxima & minima, f’(x)=0
\begin{aligned} &(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\frac{2}{3}}=0 \\ &\frac{1}{3}(x+1)=-(x+2)^{\frac{1}{3}}(x+2)^{\frac{2}{3}} \\ &\frac{1}{3}(x+1)=-(x+2) \\ &x+1=-3 x-6 \\ &x=-\frac{7}{4} \end{aligned}
At x=$-\frac{7}{4}$
\begin{aligned} f^{\prime \prime}\left(-\frac{7}{4}\right) &=\frac{2}{3}\left(-\frac{7}{4}+2\right)^{\frac{-2}{3}}-\frac{2}{9}\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\left(-\frac{5}{3}\right)} \\ &=\frac{2}{3}\left(\frac{1}{4}\right)^{-\frac{2}{3}}+\frac{1}{18}\left(\frac{1}{4}\right)^{-\frac{5}{3}}>0 \end{aligned}
So , $x=-\frac{7}{4}$ is a point of local minima
Now at $x=-\frac{7}{4}$
\begin{aligned} f\left(-\frac{7}{4}\right) &=\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\frac{1}{3}} \\ &=\left(-\frac{3}{4}\right)\left(\frac{1}{4}\right)^{\frac{1}{3}} \\ &=-\frac{3}{4^{\frac{4}{3}}} \end{aligned}
Thus, point of local minima is $-\frac{7}{4}$ & its value is $-\frac{3}{4^{\frac{4}{3}}}$

Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is -16
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5$
Explanation:
We have,
\begin{aligned} &f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5 \\ &f^{\prime}(x)=\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=-\frac{x}{\sqrt{32-x^{2}}}-\left(\frac{2 x \sqrt{32-x^{2}}+\frac{x^{3}}{\sqrt{32-x^{2}}}}{\left(32-x^{2}\right)}\right) \ldots u \sin g \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \end{aligned}
Now for local minima and local maxima
\begin{aligned} &f^{\prime}(x)=0 \\ &\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}=0 \\ &32-x^{2}=x^{2} \\ &x^{2}=16 \\ &x=\pm 4 \end{aligned}
At x=4,
\begin{aligned} f^{\prime \prime}(4) &=\frac{-4}{\sqrt{32-4^{2}}}-\left[\frac{\left(8\left(32-4^{2}\right)+4^{3}\right)}{\left(32-4^{2}\right)\left(\sqrt{32-4^{2}}\right.}\right] \\ &=-1-\frac{192}{64} \\ &=-3<0 \end{aligned}
So, x=4 is a point of local maxima.
At x=4,
\begin{aligned} f(4) &=4 \sqrt{32-(4)^{2}} \\ &=4 \sqrt{32-16} \\ &=16 \end{aligned}
At x=-4,
\begin{aligned} f^{\prime \prime}(-4) &=-\frac{4(-4)}{\sqrt{32-(-4)^{2}}}-\left[\frac{8\left(32-4^{2}\right)-(-4)^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right] \\ &=0+2=3>0 \end{aligned}
So, x=-4 is the point of local minimum.
At x=-4
\begin{aligned} f(x) &=(-4) \sqrt{32-(-4)^{2}} \\ &=-4 \sqrt{32-16}=-16 \end{aligned}
Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

## Maxima and Minima excercise 17.3 question 1 (ix)Answer:

Point of local maxima value is $\frac{a}{3}$ and it’s local maximum value is $\frac{4 a^{3}}{27}.$ Also point of local minima is a and it’s local minimum value is 0.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(\mathrm{c}_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(\mathrm{c}_{2}\right)<0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0$
Explanation:
We have,
\begin{aligned} &f(x)=x^{3}-2 a x^{2}+a^{2} x \\ &f^{\prime}(x)=3 x^{2}-4 a x+a^{2} \\ &f^{\prime \prime}(x)=6 x-4 a \end{aligned}
For maxima and minima f’(x) =0
\begin{aligned} &3 x^{2}-4 a x+a^{2}=0 \\ &x=\frac{-(-4 a) \pm \sqrt{(-4 a)^{2}-4 \times 3 \times a^{2}}}{2 \times 3} \\ &\quad=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{6} \\ &=\frac{4 a \pm 2 a}{6} \\ &x=a \text { or } x=\frac{a}{3} \end{aligned}
At x= a,
\begin{aligned} f^{\prime \prime}(a) &=6(a)-4 a \\ &=2 a>0 \end{aligned}
So, x=a is point of local minima
Now at x=a,
\begin{aligned} f(a) &=a^{3}-2 a(a)^{2}+a^{2}(a) \\ &=a^{3}-2 a^{3}+a^{3}=0 \end{aligned}
Also, at $\mathrm{x}=\frac{a}{3}$
\begin{aligned} f\left(\frac{a}{3}\right) &=\left(\frac{a}{3}\right)^{2}-2 a\left(\frac{a}{3}\right)+a^{2}\left(\frac{a}{3}\right) \\ &=\frac{4 a^{3}}{27} \end{aligned}
Thus, point of local maxima is $\frac{a}{3}$ & its max. value is $\frac{4 a^{3}}{27}$& point of local minima is a & its value is 0.

Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is -2a
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ &$c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x+\frac{a^{2}}{x}, a>0$
Explanation:
We Have,
\begin{aligned} &f(x)=x+\frac{a^{2}}{x} \\ &f^{\prime}(x)=1-\frac{a^{2}}{x^{2}} \\ &f^{\prime \prime}(x)=\frac{2 a^{2}}{x^{3}} \end{aligned}
For maxima and minima ,f’(x)=0
\begin{aligned} 1-\frac{a^{2}}{x^{2}} &=0 \\ x &=\pm a \end{aligned}

Maxima and minima exercise 17.3 question 1 (xi)

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.
Hint:
Putting x=1,-1in f''(x)
Given:
\begin{aligned} &f(x)=x \sqrt{2-x^{2}} \\ &-\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}
Explanation:
Differentiating f(x) with respect to x
$f^{\prime}(x)=\frac{d}{d x}\left(x \sqrt{2-x^{2}}\right)$
\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{g}(\mathrm{x})\}+\mathrm{g}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}+\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{-}(1) \end{aligned}
\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{df}(\mathrm{x})^{\mathrm{n}}}{\mathrm{dx}}=\mathrm{nf}(\mathrm{x})^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x}) \text { and }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})]=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x})) \pm \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{g}(\mathrm{x}))\\ &\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2-\mathrm{x}^{2}\right)^{1 / 2}\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2-1} \frac{d}{d x}\left(2-x^{2}\right)\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2}(-2 x)\\ &=-\frac{x}{\sqrt{2-x^{2}}} \end{aligned}
\begin{aligned} &\therefore \operatorname{From}(1), \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}\left(-\frac{\mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}\right)+\sqrt{2-\mathrm{x}^{2} \times 1} \\ &=\frac{-\mathrm{x}^{2}+2-\mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \\ &=\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \end{aligned}
\begin{aligned} &\text { Put, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=0\\ &\frac{2-2 x^{2}}{\sqrt{2-x^{2}}}=0\\ &\Rightarrow 2-2 x^{2}=0\\ &\Rightarrow 2 \mathrm{x}^{2}=2\\ &\Rightarrow \mathrm{x}^{2}=1\\ &\Rightarrow x=\pm 1\\ &\text { if } x \neq \sqrt{2} \text { and } x \neq-\sqrt{2} \end{aligned}
These x=1 and x=-1 we the possible point of local maxima and minima
Differentiating f’(x) with respect to x
\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}}\right) \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(2-2 \mathrm{x}^{2}\right)-\left(2-2 \mathrm{x}^{2}\right) \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}}{\left(2-\mathrm{x}^{2}\right)} \end{aligned}
Using the formula,
\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}-\mathrm{u} \frac{\mathrm{d}}{\mathrm{ax}} \mathrm{v}}{\mathrm{v}^{2}} \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \times(-4 \mathrm{x})-\frac{\left(2-2 \mathrm{x}^{2}\right) 1}{2}\left(2-\mathrm{x}^{2}\right)^{-1 / 2}(-2 \mathrm{x})}{\sqrt{2-\mathrm{x}^{2}} 2} \quad \text { (Applying chain rule) } \\ &=\frac{-4 \mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}+\frac{\mathrm{x}\left(2-2 \mathrm{x}^{2}\right)}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-4 \mathrm{x}\left(2-\mathrm{x}^{2}\right)+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-8 \mathrm{x}+4 \mathrm{x}^{3}+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \end{aligned}
\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=1 \\ &\mathrm{f}^{\prime \prime}(1)=\frac{-6(1)+2(1)^{3}}{\left(2-(1)^{2}\right)^{3 / 2}} \\ &=\frac{-6+2}{(1)^{3 / 2}}=-4<0 \end{aligned}

So x=1 is local maxima and its maximum value f(1)=1

\begin{aligned} &f^{\prime \prime}(x) \text { at } x=-1 \\ &f^{\prime \prime}(-1)=\frac{-6(-1)+(2)(-1)^{3}}{\left(2-(-1)^{2}\right)^{3 / 2}} \\ &=\frac{6-2}{(\sqrt{1})^{3}}=4>0 \end{aligned}

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

x=$\frac{3}{4}$ is a point of local maxima and its maximum value is 5/4.
Hint:
Using chain rule of Differentiation
Given:
\begin{aligned} &\mathrm{f}(\mathrm{x})=\mathrm{x}+\sqrt{1-\mathrm{x}} \\ &\mathrm{x} \leq 1 \end{aligned}
Explanation:
Differentiating f(x) with respect to x
\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\sqrt{1-\mathrm{x}}) \\ &=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{1 / 2} \\ &=1+\frac{1}{2}(1-\mathrm{x})^{\frac{1}{2}-1}(-1) \end{aligned} [ Using Chain Rule Diffrentiation ]
\begin{aligned} &=1-\frac{1}{2 \sqrt{1-x}} \\ &=\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}} \\ &\text { Put } f^{\prime}(x)=0 \\ &\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}}=0 \\ &\Rightarrow 2 \sqrt{1-x}-\mid 1=0 \text { if } x \neq 1 \\ &\Rightarrow \sqrt{1-x}=\frac{1}{2} \end{aligned} [Squaring Both Sides]
\begin{aligned} &\text { i. } e 1-x=\frac{1}{4} \\ &x=\frac{3}{4} \end{aligned}
Thus , $x=\frac{3}{4}$ is the point of local maxima and minima
Differentiating f’ with respect to x
\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(1-\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=0-\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{-1 / 2}\\ &=\frac{1}{2} \times-\frac{1}{2}(1-x)^{-3 / 2}(-1)\\ &\text { (Using chain rule) }\\ &=-\frac{1}{4(1-x)^{\frac{3}{2}}} \end{aligned}\begin{aligned} &\text { Put } \mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=\frac{3}{4} \\ &\mathrm{f}^{\prime \prime}\left(\frac{3}{4}\right)=\frac{-1}{4\left(1-\frac{3}{4}\right)^{3 / 2}} \\ &=\frac{-1}{4 \times \frac{1}{2 \sqrt{2}}} \\ &=-\frac{\sqrt{2}}{2}<0 \end{aligned}

So $x = \frac{3}{4}$is a point of local maxima
The local maximum value at f(x)
\begin{aligned} &\mathrm{f}\left(\frac{1}{2}\right)=\frac{3}{4}+\sqrt{1-\frac{3}{4}} \\ &=\frac{3}{4}+\frac{1}{2}=\frac{3+2}{4}=\frac{5}{4} \end{aligned}

## Maxima and Minima exercise 17.3 question 2 (i)

$x=2$ is the local minima and the local minimum value of $f(x)$ at $x=2$ is 0.
$x=\frac{4}{3}$ is the local maxima and the local maximum value of $f(x)$at $x=\frac{4}{3}$ is $x=\frac{4}{27}$.

Hint:
Using chain rule
Given:
$f(x)=(x-1)(x-2)^{2}$
Explanation:
Differentiating f with respect to x
\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\mathrm{x}-1)(\mathrm{x}-2)^{2}\right) \\ &=(\mathrm{x}-1) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-2)^{2}+(\mathrm{x}-2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1) \\ &=2(\mathrm{x}-1)(\mathrm{x}-2)+(\mathrm{x}-2)^{2} \quad \text { [Using chain rule] } \\ &=2\left(\mathrm{x}^{2}-2 \mathrm{x}-\mathrm{x}+2\right)+\left(\mathrm{x}^{2}-4 \mathrm{x}+4\right) \\ &=2 \mathrm{x}^{2}-4 \mathrm{x}-2 \mathrm{x}+4+\mathrm{x}^{2}-4 \mathrm{x}+4 \\ &=3 \mathrm{x}^{2}-10 \mathrm{x}+8 \end{aligned}
\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &3 x^{2}-10 x+8=0 \\ &x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{(10)^{2}-4 \cdot 3 \cdot 8}}{2 \cdot(3)} \\ &=\frac{+10 \pm \sqrt{100-96}}{6} \\ &=\frac{-10 \pm 2}{6} \end{aligned}
\begin{aligned} &\mathrm{x}=\frac{+10+2}{6}\\ &=2\\ &\text { And }\\ &x=\frac{+10-2}{6}\\ &=\frac{4}{3} \end{aligned}
Differentiating $f(x)$ with respect to $x$
\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-10 \mathrm{x}+\mathrm{i}\right) \\ &=6 \mathrm{x}-10 \end{aligned}
When put x = 2 in f’’(x)
\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}
So x=2 is the local Minima
The Local Minimum value of f(x) at x =2 is 0
Put $x=\frac{4}{3} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})$
\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}
The local maximum value of $f(x)$at $x=\frac{4}{3}$ is $x=\frac{4}{27}$.
x=1 is point of inflexion
x=-1 is the point of local minimum &$x=\frac{1}{5}$ is the point of local maximum
Hint:
Using chain rule of derivative
Given:
$f(x)=-(x-1)^{3}(x+1)^{2}$
Explanation:
Differentiating f with respect to x
\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+1)^{2}+(\mathrm{x}+1)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^{3}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \cdot 2(\mathrm{x}+1)+3(\mathrm{x}+1)^{2}(\mathrm{x}-1)^{2}\right] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)[2 \mathrm{x}-2+3 \mathrm{x}+3] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)(5 \mathrm{x}+1) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+\mathrm{x}+5 \mathrm{x}+1\right) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}+6 \mathrm{x}^{3}+\mathrm{x}^{2}-10 \mathrm{x}^{3}-12 \mathrm{x}^{2}-2 \mathrm{x}+5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}-4 \mathrm{x}^{3}-6 \mathrm{x}^{2}+4 \mathrm{x}+1\right) \end{aligned}
\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &-1(x-1)^{2}(x+1)(5 x+1)=0 \\ &(x-1)^{2}(x+1)(5 x+1)=6 \\ &\Rightarrow(x-1)^{2}=0 \\ &x-1=0 \\ &x=1 \\ &x+1=0, \quad 5 x+1=0 \\ &x=-1, \quad x=-\frac{1}{5} \end{aligned}
Thus $x=1$ and $x=-1$ and $x=-\frac{1}{5}$ are the possible points of local minima and maxima
Differentiating f’(x) with respect to x\begin{aligned} &f^{\prime \prime}(x)=\frac{d}{d x}\left[-\left(5 x^{4}-4 x^{3}-6 x^{2}+4 x+1\right)\right] \\ &=-\left(20 x^{3}-12 x^{2}-12 x+4\right) \\ &=-20 x^{3}+12 x^{2}+12 x-4 \\ &\text { when } x=1 \\ &\begin{aligned} f^{\prime \prime}(1) &=-20(1)^{3}+12(1)^{2}+12(1)-4 \\ &=-20+12+12-4=0 \end{aligned} \end{aligned}

Thus this test is fail as x=1 is point of inflexion
when $x=-1$
\begin{aligned} f^{\prime \prime}(-1) &=-20(-1)^{3}+12(-1)^{2}+12(-1)-4 \\ &=20+12-12-4=16>0 \end{aligned}

So , x = -1 is a point of local minimum

When $x=-\frac{1}{5}$

\begin{aligned} f\left(-\frac{1}{5}\right)=&-20\left(-\frac{1}{5}\right)^{3}+12\left(-\frac{1}{5}\right)^{2}+12\left(-\frac{1}{5}\right)-4 \\ &=-336 / 125<0 \end{aligned}
So, x=-1/5 is the point of local maximum

## Maxima and Minima exercise 17.3 question 3

$a = -\frac{2}{3}$
$\mathrm{b}=-\frac{1}{6}$
Hint:
Put? $\frac{d y}{d x}=0 \text { at } x=1,2$
Given:
$y=\operatorname{alog} x+b x^{2}+x$
Explanation:
Differentiating y with respect to x
\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{~b} \mathrm{x}+1 \\ &\quad=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1 \\ &\text { At } \mathrm{x}=1 \mid \\ &\begin{aligned} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) &=\frac{\mathrm{a}}{1}+2 \mathrm{~b}(1)+1 \\ &=\mathrm{a}+2 \mathrm{~b}+1-(1) \end{aligned} \end{aligned}
\begin{aligned} &\text { At } x=2 \\ &\frac{d y}{d x}=\frac{a}{2}+2(b) \cdot 2+1 \\ &=\frac{a}{2}+4 b+1-(2) \end{aligned}
Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2
Then equation (1) and (2) become
\begin{aligned} &a+2 b+1=0-(3)\\ &\frac{a}{2}+4 b+1=0\\ &a+8 b+2=0 \quad-(4)\\ &\text { [Multiplying both sides by 2] } \end{aligned}
Solving , (3) and (4) , we get
$\begin{array}{r} 6 b-1=0 \\ b=-\frac{1}{6} \end{array}$
Putting $B = -1/6 in (3) , we get$
\begin{aligned} &a-\frac{1}{3}+120 \\ &a=-\frac{2}{3} \end{aligned}

## Maxima and Minima excercise 17.3 question 4

We need to prove that x = e is local maxima
Given:
The given function $\frac{\log x}{x}$
Explanation:
$\text { Let } y=\frac{\log x}{x}$
Then
\begin{aligned} \frac{d y}{d x} &=\frac{x \cdot \frac{d}{d x}(\log x)-\log (x) \frac{d}{d x}(x)}{x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{x \times \frac{1}{x}-\log x}{x^{2}} \\ &=\frac{1-\log x}{x^{2}} \end{aligned}
Put,
\begin{aligned} &\frac{d y}{d x}=0 \\ &\frac{1-\log x}{x^{2}}=0 \\ &\Rightarrow 1-\log x=0, x \neq 0 \\ &\Rightarrow \log x=1 \\ &x=e^{1}=e \end{aligned}
Differentiating $\frac{d y}{d x}$ with respect to $x$
\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}} &=\frac{\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(1-\log \mathrm{x})-(1-\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{2}\right)}{\mathrm{x}^{4}} . \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{\mathrm{x}^{2}\left(-\frac{1}{\mathrm{x}}\right)-2 \mathrm{x}(1-\log \mathrm{x})}{\mathrm{x}^{4}} \\ &=\frac{-\mathrm{x}-2 \mathrm{x}+2 \mathrm{xlog} \mathrm{x}}{\mathrm{x}^{4}} \\ &=\frac{-3 \mathrm{x}-2 \mathrm{x} \log \mathrm{x}}{\mathrm{x}^{4}} \end{aligned}
At x=e,
\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-3 e+2 e \log e}{e^{4}} \\ &=\frac{-3 e+2 e(1)}{e^{4}} \ldots \log e=1 \\ &=-\frac{e}{e^{4}}=-\frac{1}{e^{3}}<0 \end{aligned}
So x = e is a point of local maxima

Maxima and Minima exercise 17.3 question 5

x = 0 is local minima and its minimum value is 2 and x =1 is local maxima and its maximum value is -6
Given:
$f(x)=\frac{4}{x+2}+x$
Explanation:
Differentiating f with respect to x
\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(4(\mathrm{x}+2)^{-1}+\mathrm{x}\right) \\ &=-4(\mathrm{x}+2)^{-2}+1 \\ &=-\frac{4}{(x+2)^{2}}+1 \\ &=\frac{-4+\mathrm{x}^{2}+4 \mathrm{x}+4}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}(\mathrm{x}+4)}{(\mathrm{x}+2)^{2}} \end{aligned}
\begin{aligned} &\text { Put } f(x)=0 \\ &\frac{x(x+4)}{(x+2)^{2}}=0 \\ &x(x+4)=0 \\ &\text { i.e } x=0, \text { or } \\ &x=-4 \end{aligned}
Thus x = 0 and x = -4 are the points of local maxima and minima
Differentiating f’ with respect to x
\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right) \\ &=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{(\mathrm{x}+2)^{2}(2 \mathrm{x}+4)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \cdot 2(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left[(\mathrm{x}+2)^{2}-\left(\mathrm{x}^{2}+4 \mathrm{x}\right)\right]}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left(\mathrm{x}^{2}+4 \mathrm{x}+4-\mathrm{x}^{2}-4 \mathrm{x}\right)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8}{(\mathrm{x}+2)^{3}} \end{aligned}
Put x = 0 in f ‘’ (x)
\begin{aligned} \mathrm{f}^{\prime \prime}(0) &=\frac{8}{(0+2)^{3}} \\ &=\frac{8}{8} \\ &=1>0 \end{aligned}
S,x=0is local minima. Hence,its minimum value will be f(0)=2
And x= -4 in f ‘’(x)
\begin{aligned} &f^{\prime \prime}(-4)=\frac{8}{(-4+2)^{3}} \\ &=\frac{8}{-8} \\ &=-1<0 \end{aligned}
So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6

Maxima and Minima exercise 17.3 question 6

$x=\frac{\pi}{4}$ is a point of local minima and its local minimum value will be $f\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
$x=\frac{3\pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3\pi}{4}\right)=-1-\frac{3\pi}{2}$
Hint:
$\text { Put } f^{\prime} x=0$
Given:
$f(x)=\tan x-2 x$
Explanation:
Differentiating f(x) with respect to x
\begin{aligned} &f(x)=\sec ^{2} x-2\\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0\\ &\sec ^{2} x-2=0\\ &\Rightarrow \sec ^{2} \mathrm{x}=2\\ &\Rightarrow \sec \mathrm{x}=\pm \sqrt{2}\\ &=\sec x=\sqrt{2}\\ &\Rightarrow \mathrm{x}=\frac{\pi}{4}\\ &\text { or, } \sec x=-\sqrt{2}\\ &\text { or, } \sec \mathrm{x}=\pi-\frac{\pi}{4}\\ &=\frac{3 \pi}{4} \end{aligned}
Thus $x=\frac{\pi}{4}$ or $x=\frac{3\pi}{4}$ is possible points of local maxima and minima
Differentiating f’(x) with respect to x
\begin{aligned} &\mathrm{f}^{\mathrm{u}}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec ^{2} \mathrm{x}-2\right] \\ &=2 \sec \mathrm{x}(\sec \mathrm{xtan} \mathrm{x}) \\ &=2 \sec ^{2} \mathrm{x} \tan \mathrm{x} \\ &\text { Put } \mathrm{x}=\frac{\pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x}) \\ &\begin{aligned} \mathrm{f}=\left(\frac{\pi}{4}\right) &=2 \sec ^{2}\left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{4}\right) \\ &=2(2)(1) \\ &=4>0 \end{aligned} \end{aligned}
So ,
$\mathrm{x}=\frac{\pi}{4}$ is a point of Local Minima and its local maximum value will be $\mathrm{f}\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
\begin{aligned} &\text { And Put, }\\ &\mathrm{x}=\frac{3 \pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})\\ &\mathrm{f}^{\prime \prime}\left(\frac{3 \pi}{4}\right)=2 \sec ^{2}\left(\frac{3 \pi}{4}\right) \tan \left(\frac{3 \pi}{4}\right)\\ &=2(-\sqrt{2})^{2} \quad(-1)\\ &=-4<0 \end{aligned}
So $x=\frac{3 \pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3 \pi}{4}\right)=-1-\frac{3 \pi}{2}$

Maxima and Minima exercise 17.3 question 7
\begin{aligned} &a=3 \\ &b=-9 \& c \in R \end{aligned}
Hint:
Put,
\begin{aligned} &\mathrm{f}^{\prime}(-1)=0 \\ &\mathrm{f}^{\prime}(3)=0 \end{aligned}
Given:
$f(x)=x^{3}+a x^{2}+b x+C$
Explanation:
Differentiating f(x) with respect to x
\begin{aligned} f^{\prime}(x) &=3 x^{2}+2 a x+b+0 \\ &=3 x^{2}+2 a x+b \end{aligned}
\begin{aligned} &\text { Calculate } \mathrm{f}^{\prime}(\mathrm{x}) \text { at } \mathrm{x}=-1 \text { and } \mathrm{x}=3\\ &\begin{aligned} \mathrm{f}^{\prime}(-1) &=3(-1)^{2}+2 \mathrm{a}(-1)=\mathrm{b} \\ &=3-2 \mathrm{a}+\mathrm{b} \\ \mathrm{f}^{\prime}(3) &=3(3)^{2}+2 \mathrm{a}(3)+\mathrm{b} \\ &=27+6 \mathrm{a}+\mathrm{b} \end{aligned} \end{aligned}
Put and $f^{\prime}(-1) =0$ and $f^{\prime}(3)=0$ as they are the extremum values hence $f^{\prime}(x)=0$
\begin{aligned} &\therefore 3-2 \mathrm{a}+\mathrm{b}=0 \\ &7-2 \mathrm{a}+\mathrm{b}=-3-(1) \\ &27-6 \mathrm{a}+\mathrm{b}=20 \mid \\ &6 \mathrm{a}+\mathrm{b}=-27-(2) \\ &(1)-(2), \mathrm{wc} \text { get } \\ &-2 \mathrm{a}-6 \mathrm{a}=-3+27 \\ &-8 \mathrm{a}=2 \mathrm{y} \\ &\mathrm{a}=3 \\ &\text { Put } \mathrm{a}=3 \text { in }(1) \\ &-2(3)^{2}-\mathrm{b}=-3 \\ &\mathrm{~b}=-9 \end{aligned}

## Maxima and Minima exercise 17.3 question 8

Hence proved.
Hint: $\mathrm{f}^{\prime}(\mathrm{x})=0$
Put,
Given:
$\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\sqrt{3} \cos \mathrm{x}$
Explanation:
Differentiating fx with respect to x
\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\cos x-\sqrt{3} \sin \mathrm{x} \\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0 \\ &\cos \mathrm{x}-\sqrt{3} \sin \mathrm{x}=0 \\ &\Rightarrow \sqrt{3} \sin \mathrm{x}=\cos \mathrm{x} \\ &\Rightarrow \tan \mathrm{x}=\frac{1}{\sqrt{3}} \end{aligned}
$\Rightarrow x=\frac{\pi}{6}$
Differentiating f’(x) with respect to x ,
\begin{aligned} &f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x \\ &\text { Putt } x=\frac{\pi}{6} \text { at } f^{\prime \prime}(x) \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6} \\ &=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ &=-\frac{1}{2}-\frac{3}{2} \\ &=-\frac{4}{2} \\ &=-2<0 \end{aligned}
So, $x = \frac{\pi}{6}$ is a point local maxima.

Rd Sharma Class 12th exercise 17.3 is an essential part of the RD Sharma solutions for maths. The Class 12 RD Sharma chapter 17 exercise 17.3 solution will contain an important chapter of the NCERT maths book that needs to be studied well. RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 is going to be based on the chapter Maxima and minima.
Rd Sharma class 12-chapter 17 exercises 17.3 answers that are basic and simple. It will help students learn all formulae well and become skilled at the subject in general. RD Sharma class 12 solutions ex 17.3 has around 21 inquiries.

• Solutions identified with the Greatest and least upsides of a function in its space
• Definition and which means of greatest
• Definition and significance of least
• Nearby maxima
• Questions identified with Nearby minima
• Discover Solutions of First subordinate test for neighborhood maxima and minima
• Higher-request subordinate test

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##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available
##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available
##### Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available

3 Jobs Available
##### Recreational Worker

A recreational worker is a professional who designs and leads activities to provide assistance to people to adopt a healthy lifestyle. He or she instructs physical exercises and games to have fun and improve fitness. A recreational worker may work in summer camps, fitness and recreational sports centres, nature parks, nursing care facilities, and other settings. He or she may lead crafts, sports, music, games, drama and other activities.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Big Data Analytics Engineer

Big Data Analytics Engineer Job Description: A Big Data Analytics Engineer is responsible for collecting data from various sources. He or she has to sort the organised and chaotic data to find out patterns. The role of Big Data Engineer involves converting messy information into useful data that is clean, accurate and actionable.

2 Jobs Available
##### Cloud Solution Developer

A Cloud Solutions Developer is basically a Software Engineer with specialisation in cloud computing. He or she possesses a solid understanding of cloud systems including their operations, deployment with security and efficiency with no little downtime.

2 Jobs Available
##### CRM Technology Consultant

A Customer Relationship Management Technology Consultant or CRM Technology Consultant is responsible for monitoring and providing strategy for performance improvement with logged calls, performance metrics and revenue metrics. His or her role involves accessing data for team meetings, goal setting analytics as well as reporting to executives.

2 Jobs Available
##### IT Manager

Career as IT Manager  requires managing the various aspects of an organization's information technology systems. He or she is responsible for increasing productivity and solving problems related to software and hardware. While this role is typically one of the lower-level positions within an organisation, it comes with responsibilities related to people and ownership of systems.

2 Jobs Available